Ch. 6.4 Solving Polynomial Equations. Sum and Difference of Cubes.

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Presentation transcript:

Ch. 6.4 Solving Polynomial Equations

Sum and Difference of Cubes

Factor x 3 – 125. ALGEBRA 2 LESSON 6-4 Solving Polynomial Equations x 3 – 125 = (x) 3 – (5) 3 Rewrite the expression as the difference of cubes. = (x – 5)(x 2 + 5x + (5) 2 )Factor. = (x – 5)(x 2 + 5x + 25)Simplify. 6-4

P. 322 Check understanding # 3

Solve 8x = 0. Find all complex roots. ALGEBRA 2 LESSON 6-4 Solving Polynomial Equations 8x = (2x) 3 + (5) 3 Rewrite the expression as the difference of cubes. = (2x + 5)((2x) 2 – 10x + (5) 2 )Factor. = (2x + 5)(4x 2 – 10x + 25)Simplify. The quadratic expression 4x 2 – 10x + 25 cannot be factored, so use the Quadratic Formula to solve the related quadratic equation 4x 2 – 10x + 25 = Since 2x + 5 is a factor, x = – is a root. 5252

(continued) ALGEBRA 2 LESSON 6-4 Solving Polynomial Equations x = Quadratic Formula = Substitute 4 for a, –10 for b, and 25 for c. –b ± b 2 – 4ac 2a –(–10) ± (–10) 2 – 4(4)(25) 2(4) = Use the Order of Operations. – (–10) ± – = –1 = i 10 ± 10i 3 8 = Simplify. 5 ± 5i 3 4 The solutions are – and ± 5i 3 4

P. 323 # 4 B

Factor x 4 – 6x 2 – 27. ALGEBRA 2 LESSON 6-4 Solving Polynomial Equations Step 1: Since x 4 – 6x 2 – 27 has the form of a quadratic expression, you can factor it like one. Make a temporary substitution of variables. x 4 – 6x 2 – 27 = (x 2 ) 2 – 6(x 2 ) – 27Rewrite in the form of a quadratic expression. = a 2 – 6a – 27Substitute a for x 2. Step 2: Factor a 2 – 6a – 27. a 2 – 6a – 27 = (a + 3)(a – 9) Step 3: Substitute back to the original variables. (a + 3)(a – 9) = (x 2 + 3)(x 2 – 9)Substitute x 2 for a. = (x 2 + 3)(x + 3)(x – 3)Factor completely. The factored form of x 4 – 6x 2 – 27 is (x 2 + 3)(x + 3)(x – 3). 6-4

p. 323 Check Understanding # 5B

Solve x 4 – 4x 2 – 45 = 0. ALGEBRA 2 LESSON 6-4 Solving Polynomial Equations x 4 – 4x 2 – 45 = 0 (x 2 ) 2 – 4(x 2 ) – 45 = 0 Write in the form of a quadratic expression. Think of the expression as a 2 – 4a – 45, which factors as (a – 9)(a + 5). (x 2 – 9)(x 2 + 5) = 0 (x – 3)(x + 3)(x 2 + 5) = 0 x = 3 or x = –3 or x 2 = –5Use the factor theorem. x = ± 3 or x = ± i 5Solve for x, and simplify. 6-4

Homework Page 324, #: 1, 2, 8, 9, odd